General relativity

Sommaire

1 Curved space-time
2 Riemann coordinates
3 The metric
4 Riemann tensor
5 Einstein equations in vacuum
6 Gravitational waves
7 Einstein and Newton
8 See also

Curved space-time

Curvature of the four-dimensional space-time is the basis of general relativity. A curved space is difficult to conceive particularly the fourth dimension is peculiar. Einstein calls it t=x₄. It seems simpler to consider it as an imaginary number ict where i is the quadratic root of -1 and c the speed of light. Then the space-time has the following four dimensions: (x,y,z,w=ict).

Riemann coordinates

Understanding of general relativity, like restricted relativity, will be easier by using two dimensions (x, y=ict) instead of four. With this representation, we will have a riemannian instead of pseudo-riemannian space. Cartesian coordinates are the most common reference system. The Earth, being spherical, is not a flat space and the Pythagorean theorem is valid only locally. The cartesian frame changes its orientation from place to place but the law of gravity is the same in Paris or in Valparaiso. The Riemann coordinates are local cartesian coordinates. They are such that the Pythagorean theorem is valid even on a curved surface. It is not necessary to know the transformation from curved coordinates to use them. They are not always suitable, for example, it is necessary to compute the Riemann tensor in Gauss (e.g. spherical) coordinates in order to obtain the Schwarzschild metric.

The metric

The metric of a euclidean space represents, in the plane, the Pythagorean theorem.

$LaTeX: \left. ds^2= dx^2 + dy^2\right.$

The metric of a curved surface is, according to Gauss:

$LaTeX: \left. ds^2= g_{xx} dx^2 + 2 g_{xy} dx dy + g_{yy} dy^2\right.$

where the g_ij are the coefficients of the metric. Every curved surface may be approximated, locally, by the osculating paraboloid, becoming the tangent plane z=0 when the principal curvatures k_x et k_y cancel:

$LaTeX: z= \frac{1}{2}\left(k_{x} x^2 + k_{y} y^2\right)$

Indeed, in the frame used, the axes Ox and Oy are in the tangent plane z=0, the origin of the coordinates, x=0, y=0 being at the contact point. The Gauss curvature is, by definition, the product of the principal curvatures:

$LaTeX: K=k_{x}k_{y}= \frac{\partial ^2 z}{\partial x^2} \frac{\partial ^2 z}{\partial y^2}$

In order to be in Riemann coordinates, it remains to orientate the axes Ox and Oy in such a manner that the metric be diagonal (the computation is given in <ref name="lire1">Bernard Schaeffer, Relativités et quanta clarifiés, Publibook, 2007</ref>):

$LaTeX: ds^2= dx^2 + \left[1 - K\left( x^2 + y^2\right)\right] dy^2$

where K= k_xk_y is the Gaussian curvature. In this expression, we have g_xx=1, g_xy=0 and

$LaTeX: g_{yy} =1 - K\left( x^2 + y^2\right)$

It is not necessary to determine the principal directions to work with the Riemann coordinates since the laws of physics are invariant under a frame change. It is also not necessary to change the scales of the coordinate axes to get a metric with coefficients equal to one. It only assumed that it is always possible to change the coordinates in such a way that the Pythagorean theorem is verified locally, at the contact point, taken as the origin of the coordinates. In Riemann coordinates, all the paraboloids, including the sphere, locally, have the same metric, provided thet have the same Gaussian curvature.

Riemann tensor

Gauss found a formula of the curvature K of a surface with a computation, complicated in Gaussian coordinates but much simpler in Riemannian coordinates where the curvature and the Riemann tensor are equal (in two dimensions):

$LaTeX: R_{xyxy}= -\frac{1}{2}\left(\frac{\partial ^2 g_{xx}}{\partial y^2} + \frac{\partial ^2 g_{yy}}{\partial x^2}\right)$

Let us check that the Riemann tensor is equal to the total Gauss curvature:

$LaTeX: R_{xyxy}= -\frac{1}{2}\left(\frac{\partial ^2 g_{xx}}{\partial y^2} + \frac{\partial ^2 g_{yy}}{\partial x^2}\right)= 0 -\frac{1}{2}(-2K)=K$

We have also, by partial derivation of the coefficients of the metric:

$LaTeX: \frac{\partial ^2 g_{xx}}{\partial x^2} + \frac{\partial ^2 g_{xx}}{\partial y^2}= 0$

The same for g_yy

$LaTeX: \frac{\partial ^2 g_{yy}}{\partial x^2} + \frac{\partial ^2 g_{yy}}{\partial y^2}= -4K$

We have obtained a Laplace equation and a Poisson equation.

Einstein equations in vacuum

Einstein's hypothesis is that the curvature of space-time is zero in the vacuum which is thus a flat space. This is true in two dimensions where the Gaussian curvature is zero. In higher dimensions, only the Ricci tensor is zero according to the Einstein equation. In matter, the Ricci tensor is different from zero. We shall not consider this case, here, but it should be considered to describe the universe which contains matter. The Einstein equations are, in the vacuum:

$LaTeX: \left.R_{ik}= 0\right.$

R_ik is a complicated function of the various componants of the Riemann tensorR_ijkl and of the metric g_ik. The Ricci tensor, like the Riemann tensor dépends only on the coefficients of the metric. The Christoffel symbols http://en.wikipedia.org/wiki/Christoffel_symbol are then unnecessary intermediaries. In two dimensions, the Ricci tensor has two components each proportional to the single component of the Riemann tensor. Therefore there is only one Einstein equation in two dimensions:

$LaTeX: \left.R_{xyxy}= 0\right.$

In two dimensions and in Riemann coordinates, the Riemann tensor is equal to the Gaussian curvature K, which is zero in the vacuum. Then the coefficients of the metric have to satisfy the Laplace equation Δg_xx=0 and Δg_yy=0. But, in two dimensions, the Laplace equation diverges unless the coefficients of the metric are constants, corresponding to a pseudo-euclidean space. In three and four dimensions, the Ricci tensor has to be zero, the corresponding space is called Ricci flat. The calculation is too complicated to be given here.

Gravitational waves

Replacing y by ict in the Laplace equation, one obtains the d'Alembert equation of the plane gravitational waves for the coefficients of the metric:

$LaTeX: \frac{\partial ^2 g_{xx}}{\partial x^2} - \frac{1}{c^2}\frac{\partial ^2 g_{xx}}{\partial t^2} = 0$

$LaTeX: \frac{\partial ^2 g_{tt}}{\partial x^2} - \frac{1}{c^2}\frac{\partial ^2 g_{tt}}{\partial t^2} = 0$

The gravitational waves have not yet been detected.

Einstein and Newton

The two-dimensional Laplace equation may be extrapolated in higher spaces with small curvature. In three dimensions, spherical symmetry and time independent metric, the Einstein equations reduce to the radial laplacian:

$LaTeX: \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dg_{rr}}{dr}\right)= 0$

and for $LaTeX: g_{tt}$ . Its solution is the Coulomb potential in 1/r:

$LaTeX: ds^2= g_{rr} dr^2 - g_{tt} c^2 dt^2= \left(A+ \frac{B}{r}\right) dr^2 -c^2 \left(A'+ \frac{B'}{r}\right) dt^2$

The correspondence principle with special relativity will give us the integration constants A and A'. For r=∞, we have:

$LaTeX: ds^2= \left.A dr^2 -c^2 A' dt^2\right.$

It should be the Minkowski metric:

$LaTeX: ds^2=\left.dr^2-c^2dt^2\right.$

Identifying these two metrics, we get A=A'=1. To obtain B', we apply the correspondence principle with the newtonian gravitation of a light particle on a circular trajectory around a highly attracting star similar to a black hole. Then dr=0, the metric is simplified:

$LaTeX: \left. ds^2= -c^2g_{tt} dt^2 = -c^2\left( 1+ \frac{B'}{R}\right) dt^2\right.$

The Minkowski metric may be written

$LaTeX: ds^2=-\left(1 -\frac{v^2}{c^2}\right)dt^2$

where v=dr/dt is the velocity of the particle. For a photon, v=c, ds=0: the length of a light trajectory is zero. It is the shortest way possible. Assuming that this remains true in general relativity, we have the condition:

$LaTeX: 1+ \frac{B'}{R}=0$

which gives R=-B'. The trajectory being a circle and the curvature of space small, we may apply newtonian mechanics. The kinetic energy is equal to the newtonian gravitation potential:

$LaTeX: \frac{1}{2}mv^2=\frac{GM}{R}$

where G is the gravitation constant, M the mass of the attracting star and c the speed of light. Replacing R with -B' and v with c, we get:

$LaTeX: B'= -\frac{2GM}{c^2}$

According to Einstein, the determinant (or its trace for low gravitation) of the metric should be equal to one. This can be shown by solving the four-dimensional Einstein equations for a static and spherically symmetric gravitational field. Therefore we may write B=-B' and obtain an approximation of the Schwarzschid metric:

$LaTeX: \left. ds^2= \left(1+ \frac{2GM}{c^2r}\right) dr^2 -c^2 \left(1- \frac{2GM}{c^2r}\right) dt^2\right.$

This metric gives a light deviation by the sun twice as predicted by the newtonian theory or by the first Einstein theory of 1911 where time is dilated by gravitation. In his 1916 theory, gravitation dilates time and contracts space.

General relativity

Sommaire

Curved space-time

Riemann coordinates

The metric

Riemann tensor

Einstein equations in vacuum

Gravitational waves

Einstein and Newton

See also

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